CBSE Class 10 Science Question Paper 2025 Solved Code 31-1-1

SECTION A

1. Electrolysis of water is a decomposition reaction. The mass ratio \( (M_H : M_O) \) of hydrogen and oxygen gases liberated at the electrodes during electrolysis of water is: [1 Mark]
(A) 8 : 1
(B) 2 : 1
(C) 1 : 2
(D) 1 : 8

Answer: (D) 1 : 8

Teacher's Note:
a) Water gives hydrogen and oxygen in a mole ratio of 2:1, but mass ratio is different because oxygen is heavier.
b) Use molar masses: 2 moles H\(_2\) = 4 g, 1 mole O\(_2\) = 32 g, so mass ratio is 4:32 = 1:8.

2. The products formed when Aluminium and Magnesium are burnt in the presence of air respectively are: [1 Mark]
(A) Al₃O₄ and MgO₂
(B) Al₂O₃ and MgO
(C) Al₃O₄ and MgO
(D) Al₂O₃ and MgO₂

Answer: (B) Al₂O₃ and MgO

Teacher's Note:
a) Balance the valency of the metal with oxygen to get the correct formula.
b) Aluminium has valency 3, so it forms Al\(_2\)O\(_3\); Magnesium has valency 2, so it forms MgO.

3. The following table shows the pH values of four solutions A, B, C and D on a pH scale: [1 Mark]
pH scale: 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14
The solutions A, B, C and D respectively are of a (A) Strong acid, weak acid, neutral, strong base
(B) Weak acid, neutral, weak base, strong base
(C) Weak acid, neutral, strong base, weak base
(D) Weak acid, neutral, strong base, strong acid

[Figure: pH scale from 1 to 14 with D marked at pH 3, A marked at pH 5, B marked at pH 7, and C marked at pH 13]

Answer: (D) Weak acid, neutral, strong base, strong acid

Teacher's Note:
a) pH less than 7 is acidic, pH 7 is neutral, and pH more than 7 is basic.
b) A pH value very close to 0 or 14 shows a strong acid or strong base; values closer to 7 show weak acid or weak base.

4. Consider the following reactions:
(i) Dilute hydrochloric acid reacts with sodium hydroxide.
(ii) Magnesium oxide reacts with dilute hydrochloric acid.
(iii) Carbon dioxide reacts with sodium hydroxide.
It is found that in each case: [1 Mark]
(A) Salt and water is formed.
(B) Neutral salts are formed.
(C) Hydrogen gas is formed.
(D) Acidic salts are formed.

Answer: (A) Salt and water is formed.

Teacher's Note:
a) All three are neutralisation type reactions between an acid and a base or a basic oxide.
b) Note that CO\(_2\) is an acidic oxide, so it also gives salt and water with a base.

5. Reaction between two elements A and B, forms a compound C. A loses electrons and B gains electrons. Which one of the following properties will not be shown by compound C? [1 Mark]
(A) It has high melting point.
(B) It is highly soluble in water.
(C) It has weak electrostatic forces of attraction between its oppositely charged ions.
(D) It conducts electricity in its molten state or aqueous solution.

Answer: (C) It has weak electrostatic forces of attraction between its oppositely charged ions.

Teacher's Note:
a) Compound C is an ionic compound since electrons transfer between A and B.
b) Ionic compounds have strong electrostatic forces, not weak ones, hence option (C) is false.

6. The metals obtained from their molten chlorides by the process of electrolytic reduction are: [1 Mark]
(A) Gold and silver
(B) Calcium and magnesium
(C) Aluminium and silver
(D) Sodium and iron

Answer: (B) Calcium and Magnesium

Teacher's Note:
a) Very reactive metals at the top of the reactivity series are extracted by electrolysis of their molten chlorides.
b) Gold and silver are unreactive; aluminium is extracted from molten oxide (alumina), not chloride.

7. The formation of magnesium oxide is correctly shown in option: [1 Mark]

[Figure: Four options (A)-(D) showing electron dot diagrams of Mg reacting with O to form MgO. Option (A) shows Mg transferring 2 electrons to one O atom to form Mg\(^{2+}\)[O]\(^{2-}\); Option (B) shows transfer of 1 electron forming Mg\(^{+}\)[O]\(^{-}\); Option (C) shows Mg transferring electrons to two O atoms forming Mg\(^{2+}\)[O]\(^{-}\)\(_2\); Option (D) shows 2Mg with O\(_2\) forming [Mg\(^{2+}\)]\(_2\)[O]\(^{2-}\)]

Answer: (A), showing Mg losing 2 electrons which are gained by one oxygen atom to form Mg\(^{2+}\)[O]\(^{2-}\)

Teacher's Note:
a) Magnesium has 2 valence electrons and oxygen needs 2 electrons to complete its octet, so one Mg atom transfers both electrons to one O atom.
b) The correct product formula must be MgO, with charges balancing to zero.

8. Secretion of less saliva in mouth will effect the conversion of: [1 Mark]
(A) proteins into amino acids
(B) fats into fatty acids and glycerol
(C) starch into simple sugars
(D) sugars into alcohol

Answer: (C) starch into simple sugars

Teacher's Note:
a) Saliva contains salivary amylase which breaks down starch into simpler sugars.
b) Protein and fat digestion occur later in the stomach and small intestine, not by saliva.

9. The plant hormone whose concentration stimulates the cells to grow longer on the side of the shoot which is away from light is: [1 Mark]
(A) Cytokinins
(B) Gibberellins
(C) Adrenaline
(D) Auxins

Answer: (D) Auxins

Teacher's Note:
a) Auxin accumulates on the shaded side of the shoot, causing more elongation there and bending towards light.
b) Adrenaline is an animal hormone, not a plant hormone, so it can be eliminated immediately.

10. The correct/true statement(s) for a bisexual flower is/are: [1 Mark]
(i) They possess both stamen and pistil.
(ii) They possess either stamen or pistil.
(iii) They exhibit either self-pollination or cross-pollination.
(iv) They cannot produce fruits on their own.
(A) (i) only
(B) (iv) only
(C) (i) and (iii)
(D) (i) and (iv)

Answer: (C) (i) and (iii)

Teacher's Note:
a) A bisexual flower has both male (stamen) and female (pistil) parts.
b) Such flowers can undergo self or cross pollination and can produce fruit on their own.

11. If pea plants with round and green seeds (RRyy) are crossed with pea plants having wrinkled and yellow seeds (rrYY), the seeds developed by the plants of F₁ generation will be: [1 Mark]
(A) 50% round and green
(B) 75% wrinkled and green
(C) 100% round and yellow
(D) 75% wrinkled and yellow

Answer: (C) 100% round and yellow

Teacher's Note:
a) Round (R) and yellow (Y) are dominant traits, so all F\(_1\) offspring (RrYy) show the dominant phenotype.
b) The seeds we see are actually part of the next generation, but here their genotype is RrYy giving round and yellow appearance.

12. The breakdown of glucose has taken the following pathway:
Glucose \(\xrightarrow{(a)}\) Pyruvate + Energy \(\xrightarrow{(b)}\) Lactic acid + Energy
The sites 'a' and 'b' respectively are: [1 Mark]
(A) Mitochondria and Oxygen deficient muscle cells
(B) Cytoplasm and Oxygen rich muscle cells
(C) Cytoplasm and Yeast cells
(D) Cytoplasm and Oxygen deficient muscle cells

Answer: (D) Cytoplasm and Oxygen deficient muscle cells

Teacher's Note:
a) Glycolysis (glucose to pyruvate) always occurs in the cytoplasm.
b) Pyruvate converts to lactic acid in muscle cells only when oxygen supply is insufficient.

13. Mirror 'X' is used to concentrate sunlight in solar furnace and Mirror 'Y' is fitted on the side of the vehicle to see the traffic behind the driver. Which of the following statements are true for the two mirrors? [1 Mark]
(i) The image formed by mirror 'X' is real, diminished and at its focus.
(ii) The image formed by mirror 'Y' is virtual, diminished and erect.
(iii) The image formed by mirror 'X' is virtual, diminished and erect.
(iv) The image formed by mirror 'Y' is real, diminished and at its focus.
(A) (i) and (ii)
(B) (ii) and (iii)
(C) (iii) and (iv)
(D) (i) and (iv)

Answer: (A) (i) and (ii)

Teacher's Note:
a) Mirror X used for solar furnaces is a concave mirror, which forms a real image at focus for parallel rays.
b) Mirror Y used for vehicles is a convex mirror, which always forms a virtual, erect, diminished image.

14. An old person is suffering from an eye defect caused by weakening of ciliary muscles and diminishing flexibility of the eye lens. If the defect of vision is 'a' which can be corrected by lens 'b', then 'a' and 'b' respectively are: [1 Mark]
(A) hypermetropia and convex lens
(B) presbyopia and bifocal lens
(C) myopia and concave lens
(D) myopia and bifocal lens

Answer: (B) presbyopia and bifocal lens

Teacher's Note:
a) Presbyopia occurs due to ageing, when ciliary muscles weaken and lens loses flexibility.
b) Bifocal lenses correct both near and distant vision in presbyopia.

15. Which of the following groups do not constitute a food chain? [1 Mark]
(i) Wolf, rabbit, grass, lion
(ii) Plankton, man, grasshopper, fish
(iii) Hawk, grass, snake, grasshopper, frog
(iv) Grass, snake, wolf, tiger
(A) (i) and (iv)
(B) (i) and (iii)
(C) (ii) and (iii)
(D) (ii) and (iv)

Answer: (D) (ii) and (iv)

Teacher's Note:
a) A valid food chain must show a correct sequence where each organism is eaten by the next one in order.
b) In group (ii) and (iv), the order of organisms does not follow a proper feeding sequence.

16. The percentage of solar energy which is not converted into food energy by the leaves of green plants in a terrestrial ecosystem is about: [1 Mark]
(A) 1%
(B) 10%
(C) 90%
(D) 99%

Answer: (D) 99%

Teacher's Note:
a) Only about 1% of the sunlight falling on leaves is captured and converted into food energy.
b) Remember this is different from the 10% rule of energy transfer between trophic levels.

17. Assertion (A): Decomposition reactions are generally endothermic reactions.
Reason (R): Decomposition of organic matter into compost is an exothermic process. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).

Teacher's Note:
a) Both statements are individually correct facts about decomposition reactions.
b) The reason given does not explain why decomposition reactions in general are endothermic, so it is not a correct explanation.

18. Assertion (A): A human child bears all the basic features of human beings.
Reason (R): It looks exactly like its parents, showing very little variations. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (C) Assertion (A) is true, but Reason (R) is false.

Teacher's Note:
a) A child does have all basic human features due to inheritance of DNA from parents.
b) The child does not look exactly like parents; sexual reproduction always introduces variations, so the reason is false.

19. Assertion (A): No two magnetic field lines are found to cross each other.
Reason (R): The compass needle cannot point towards two directions at the point of intersection of two magnetic field lines. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Teacher's Note:
a) At any point, the field line gives the unique direction of the magnetic field at that point.
b) If two lines crossed, the compass needle would need to point in two directions at once, which is impossible.

20. Assertion (A): The amount of ozone in the atmosphere began to drop sharply in the 1980s.
Reason (R): The oxygen atoms combine with molecular oxygen to form ozone. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).

Teacher's Note:
a) Both statements are factually correct about the ozone layer.
b) The drop in ozone in the 1980s was due to CFCs and other pollutants depleting ozone, not because of the way ozone naturally forms.

SECTION B

21. A student performs the following experiment in his school laboratory. List two observations to justify that in this experiment a chemical change has taken place. [2 Marks]

[Figure: A conical flask fitted with a cork and delivery tube, containing dilute sulphuric acid and zinc granules]

Answer:
1. A gas is evolved with bubbling in the flask, which is hydrogen gas.
2. There is a rise in temperature of the reaction mixture, showing an exothermic chemical change.

Teacher's Note:
a) Gas evolution and temperature change are both valid signs of a chemical reaction.
b) Zinc displaces hydrogen from dilute sulphuric acid to form zinc sulphate and hydrogen gas.

22. Draw labelled diagrams to show different stages of budding in Hydra. [2 Marks]

Answer: The diagram shows three stages: first, a small bud arises as an outgrowth on the body of the parent Hydra; second, the bud enlarges and develops tentacles while remaining attached to the parent; third, the fully formed new individual detaches from the parent to become an independent organism. Labels used are: parent Hydra, bud, tentacles, and detached new Hydra.

Teacher's Note:
a) Marks are given for both a correctly drawn outgrowth (bud) and clear labelling of stages.
b) Budding is a type of asexual reproduction seen in Hydra.

23. (a) Besides minimising the loss of blood, why is it essential to plug any leak in a blood vessel? Name the component of blood which helps in this process and state how this component perform this function. [2 Marks]

Answer:
1. Plugging the leak prevents the fall in blood pressure and maintains the efficiency of the pumping system of the heart.
2. Platelets help in this process by clotting the blood at the site of injury.

Teacher's Note:
a) Remember platelets, not red blood cells, cause clotting.
b) Clot formation prevents excessive blood loss and infection at the wound site.

OR

(b) (i) Plants have a transport system that is relatively slower than in animals. Give reasons.
(ii) State the role of phloem in the transport of materials in plants. [2 Marks]

Answer:
1. Plants have low energy needs since a large proportion of their body consists of dead cells in tissues, and they do not move, so slow transport is enough.
2. Phloem transports soluble products of photosynthesis, like sugars, from leaves to other parts of the plant, including storage organs like roots, fruits and seeds.

Teacher's Note:
a) Xylem transports water and minerals, phloem transports food; do not mix up the two.
b) Phloem transport is called translocation and can occur in both upward and downward directions.

24. An object is placed at a distance of 60 cm from a concave lens of focal length 30 cm. Use lens formula to find the position of the image formed in this case. [2 Marks]

Answer:
1. Given: u = - 60 cm, f = - 30 cm
2. Lens formula: \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)
3. \( \frac{1}{-30} = \frac{1}{v} - \frac{1}{-60} \)
4. \( \frac{1}{v} = \frac{1}{-30} - \frac{1}{60} = \frac{-3}{60} \)
5. v = - 20 cm
6. The image is formed at 20 cm from the concave lens, on the same side as the object.

Teacher's Note:
a) Always use correct sign convention: distances measured against incident light are negative.
b) A concave lens always forms a virtual, erect, diminished image on the same side as the object.

25. (a) A wire of resistance R is cut into three equal parts. If these three parts are then joined in parallel, calculate the total resistance of the combination so formed. [2 Marks]

Answer:
1. Resistance of each part = \( \frac{R}{3} \)
2. For parallel combination: \( \frac{1}{R_p} = \frac{1}{R/3} + \frac{1}{R/3} + \frac{1}{R/3} = \frac{9}{R} \)
3. Therefore, \( R_p = \frac{R}{9} \)

Teacher's Note:
a) Cutting a wire reduces its length, so resistance of each part becomes R/3, not R.
b) The final answer R/9 shows how parallel combination reduces total resistance greatly.

OR

(b) Define electric power. When do we say that the power consumed in an electric circuit is 1 watt? [2 Marks]

Answer:
1. Electric power is the rate at which electrical energy is consumed or dissipated in an electric circuit.
2. The power consumed is said to be 1 watt when a current of 1 ampere flows through a circuit at a potential difference of 1 volt.

Teacher's Note:
a) Power formula: \( P = VI \), SI unit is watt.
b) 1 watt also equals 1 joule of energy consumed per second.

26. "Excessive use of chemicals and pesticides in agriculture adversely effect the environment." Justify this statement. [2 Marks]

Answer:
1. Chemicals and pesticides used by farmers get washed into soil and water bodies, affecting both biotic and abiotic components of the ecosystem.
2. These chemicals are mostly non-biodegradable and accumulate progressively at each trophic level through biological magnification, harming organisms at all levels.

Teacher's Note:
a) The key term expected here is "biological magnification".
b) Higher trophic level organisms, including humans, get the highest concentration of these harmful chemicals.

SECTION C

27. (a) "Displacement reactions also play a key role in extracting metals in the middle of the reactivity series." Justify this statement with two examples.
(b) Why can metals high up in the reactivity series not be obtained by reduction of their oxides by carbon? [3 Marks]

Answer:
1. \( 3MnO_2(s) + 4Al(s) \rightarrow 3Mn(l) + 2Al_2O_3(s) + heat \)
2. \( Fe_2O_3(s) + 2Al(s) \rightarrow 2Fe(l) + Al_2O_3(s) + heat \) - in both cases, aluminium displaces the metal from its oxide, showing displacement reactions extract metals of medium reactivity.
3. Metals high in the reactivity series like Na, Mg and Ca have a much greater affinity for oxygen than carbon, so carbon cannot reduce their oxides.

Teacher's Note:
a) These reactions are called thermit reactions and are highly exothermic.
b) Remember the key reasoning: affinity for oxygen decides whether carbon can act as a reducing agent.

28. (a) With the help of an activity, explain the conditions under which iron articles get rusted. [3 Marks]

Answer:
1. Take three test tubes A, B and C, and place a clean iron nail in each.
2. In test tube A, pour some ordinary water and cork it; in test tube B, pour boiled distilled water, add about 1 mL of oil on top, and cork it, so air cannot dissolve in the water; in test tube C, put some anhydrous calcium chloride (which absorbs moisture) and cork it.
3. After a few days, the nail in test tube A rusts, but nails in B and C do not rust.
4. This shows that rusting of iron occurs only when the iron is exposed to both air and water together.

Teacher's Note:
a) Test tube B removes dissolved air, and test tube C removes moisture, isolating each factor.
b) A neat labelled diagram of the three test tubes can also be given instead of description.

OR

(b) (i) Name two metals which react violently with cold water. List any three observations which a student notes when these metal are dropped in a beaker containing water.
(ii) Write a test to identify the gas evolved (if any) during the reaction of these metals with water. [3 Marks]

Answer:
1. Sodium and potassium react violently with cold water (any two of sodium, potassium, lithium accepted).
2. Observations: the reaction is very vigorous; a large amount of heat is evolved; the evolved gas may catch fire on the surface of water.
3. Test: bring a burning matchstick near the gas; it burns with a pop sound, confirming the gas is hydrogen.

Teacher's Note:
a) Sodium and potassium float and move about rapidly on water due to heat produced.
b) The pop sound test is the standard way to confirm hydrogen gas in exams.

29. Plants have neither a nervous system nor muscles, even then they respond to stimuli. For example, the leaves of chhui-mui (touch-me-not) plant when touched begin to fold up and droop.
(a) How is the information communicated in "touch-me-not" plants?
(b) What enables the plant cells to bring out the observable response?
(c) Differentiate the movement mentioned above from the movement of tendrils in a pea plant. [3 Marks]

Answer:
1. Plant cells use electrical-chemical signals to convey the information that touch has occurred, spreading the message from cell to cell without any nerves.
2. Plant cells change shape by altering the amount of water in them, either swelling or shrinking, which brings about the observable movement.
3. The touch-me-not movement is independent of growth and does not occur in the direction of the stimulus (it is a nastic movement), while the tendril movement of the pea plant is growth-dependent and occurs in the direction of the stimulus (it is a tropic movement, specifically thigmotropism).

Teacher's Note:
a) Remember the key terms "nastic movement" and "tropic movement" for full marks.
b) Touch-me-not response is fast because it does not depend on growth, unlike tendril coiling.

30. (a) What are chromosomes?
(b) Explain in brief how stability of DNA content of a species is ensured in sexually reproducing organisms? [3 Marks]

Answer:
1. Chromosomes are thread-like structures in the nucleus that carry genes in the form of DNA, which control the inherited traits of an organism.
2. Each cell normally has two copies of every chromosome, one inherited from each parent; during formation of germ cells (gametes), this number is halved.
3. When two germ cells fuse during fertilisation, the zygote formed restores the normal number of chromosomes and the same amount of DNA as in the parents, keeping the DNA content stable across generations.

Teacher's Note:
a) Halving of chromosome number happens during gamete formation (meiosis).
b) Doubling back to normal number happens at fertilisation, which is the key to DNA stability.

31. Draw ray diagrams to show the nature, position and relative size of the image formed by a convex mirror when the object is placed (i) at infinity and (ii) between infinity and pole P of the mirror. [3 Marks]

Answer: (i) When the object is at infinity, parallel rays fall on the convex mirror and appear to diverge from the focus F behind the mirror after reflection; the image formed is virtual, erect, highly diminished, and located at the focus. (ii) When the object is placed between infinity and the pole P, the reflected rays appear to diverge from a point between the pole and focus behind the mirror; the image formed is virtual, erect and diminished, located between the pole and focus.

Teacher's Note:
a) In both cases for a convex mirror, the image is always virtual, erect and smaller than the object.
b) Marks are lost if arrows showing direction of light rays are not marked on the diagram.

32. Consider the following electric circuit:
Calculate the values of the following: [3 Marks]
(a) The total resistance of the circuit
(b) The total current drawn from the source
(c) Potential difference across the parallel combination of 10 \( \Omega \) and 15 \( \Omega \) resistors

[Figure: A circuit with a 15V battery connected through a key K in series with two parallel combinations: a 10 \( \Omega \) resistor parallel with a 15 \( \Omega \) resistor, in series with a 60 \( \Omega \) resistor parallel with a 40 \( \Omega \) resistor, and an ammeter A connected in the circuit]

Answer:
1. For 10 \( \Omega \) and 15 \( \Omega \) in parallel: \( \frac{1}{R_1} = \frac{1}{10} + \frac{1}{15} = \frac{1}{6} \), so \( R_1 = 6\,\Omega \)
2. For 60 \( \Omega \) and 40 \( \Omega \) in parallel: \( \frac{1}{R_2} = \frac{1}{60} + \frac{1}{40} \), so \( R_2 = 24\,\Omega \)
3. Since \( R_1 \) and \( R_2 \) are in series: \( R_{total} = 6 + 24 = 30\,\Omega \)
4. Total current: \( I = \frac{V}{R} = \frac{15}{30} = 0.5\,A \)
5. Potential difference across parallel combination of 10 \( \Omega \) and 15 \( \Omega \): \( V = I \times R_1 = 0.5 \times 6 = 3.0\,V \)

Teacher's Note:
a) First reduce each parallel group to a single equivalent resistance before adding them in series.
b) Use Ohm's law carefully at each stage, applying the correct resistance and current values.

33. (a) Write the relationship between resistivity and resistance of a cylindrical conductor of length l and area of cross-section A. Hence derive the SI unit of resistivity.
(b) Why are alloys used in electrical heating devices? [3 Marks]

Answer:
1. Resistance is related to resistivity by \( R = \rho \frac{l}{A} \), so \( \rho = \frac{RA}{l} \)
2. SI unit of resistivity = \( \frac{ohm \times m^2}{m} \) = ohm metre (\( \Omega m \))
3. Alloys are used in heating devices because their resistivity is higher than that of pure metals and they do not oxidise (burn) easily even at high temperatures.

Teacher's Note:
a) Deriving the unit correctly by cancelling out length terms is important for full marks.
b) Nichrome is a common example of an alloy used in heating elements.

SECTION D

34. (A) (i) Draw two isomeric structures of Butene (C₄H₈).
(ii) Name the following compounds:
(I) H-C-C-C-Cl with H atoms attached (a straight chain of 3 carbons with Cl on the terminal carbon)
(II) H-C-C-C-H with H, H, O double bond arrangement (a 4-carbon chain with a C=O group on the third carbon)
(iii) Write the chemical equations for the following reactions. Mention one essential condition each for these reactions to take place.
(I) Ethanol undergoes complete oxidation
(II) Propene undergoes hydrogenation
(III) Ethanoic acid reacts with ethanol [5 Marks]

[Figure: Structural formulas showing (I) 1-chloropropane, CH₃-CH₂-CH₂-Cl and (II) butanone, CH₃-CH₂-CO-CH₃]

Answer:
1. Two isomers of butene: CH\(_3\)-CH\(_2\)-CH=CH\(_2\) (but-1-ene) and CH\(_3\)-CH=CH-CH\(_3\) (but-2-ene); a third isomer, 2-methylpropene, is also acceptable.
2. (I) Chloropropane; (II) Butanone (butan-2-one)
3. Complete oxidation of ethanol: \( CH_3CH_2OH \xrightarrow{alkaline\ KMnO_4,\ heat} CH_3COOH \)
4. Hydrogenation of propene: \( CH_3CH{=}CH_2 + H_2 \xrightarrow{Ni/Pd} CH_3CH_2CH_3 \)
5. Esterification: \( CH_3COOH + C_2H_5OH \xrightarrow{acid} CH_3COOC_2H_5 + H_2O \)

Teacher's Note:
a) Isomers of butene must have the same molecular formula \( C_4H_8 \) but different structures.
b) The condition for each reaction (oxidising agent, catalyst, or acid) is essential for full marks.
c) The sweet smelling ester formed in reaction (III) is a key identifying feature of esterification.

OR

(B) (i) A carbon compound X is a good solvent. On reaction with sodium, X forms two products Y and Z. Z is used to convert vegetable oil into vegetable ghee. Identify and name X, Y and Z. Also write the equation of reaction of X with sodium to justify your answer.
(ii) Write chemical equation to show what happens when ethanol:
(I) burns in oxygen/air.
(II) is heated at 443 K in excess conc. H₂SO₄.
(III) reacts with acidified potassium dichromate. [5 Marks]

Answer:
1. X = Ethanol (\( C_2H_5OH \)); Y = Sodium ethoxide (\( C_2H_5ONa \)); Z = Hydrogen (\( H_2 \))
2. \( CH_3CH_2OH + Na \rightarrow CH_3CH_2ONa + \frac{1}{2}H_2 \)
3. Combustion: \( 2C_2H_5OH + 7O_2 \rightarrow 4CO_2 + 6H_2O + Heat + Light \)
4. Dehydration: \( C_2H_5OH \xrightarrow[Conc.\ H_2SO_4]{443\,K} C_2H_4 + H_2O \)
5. Oxidation: \( C_2H_5OH \xrightarrow{Acidified\ K_2Cr_2O_7} CH_3COOH \)

Teacher's Note:
a) Hydrogen gas (Z) is used in hydrogenation of vegetable oils to make vegetable ghee.
b) At 443K with excess sulphuric acid, ethanol undergoes dehydration to form ethene, not oxidation.
c) Acidified potassium dichromate is an oxidising agent that converts ethanol to ethanoic acid.

35. (A) (i) Write the functions of the following parts of human female reproductive system:
(I) Ovary
(II) Fallopian tube
(III) Uterus
(ii) State briefly two contraceptive methods used by human males. [5 Marks]

Answer:
1. Ovary: Produces the female gamete (egg/ovum) and secretes the female hormone oestrogen.
2. Fallopian tube: This is the site where fertilisation of the egg by the sperm takes place.
3. Uterus: This is the site of implantation of the embryo and its subsequent development.
4. Mechanical barrier method: Condoms are used by males to prevent sperm from reaching the egg.
5. Surgical method: Vasectomy involves blocking the vas deferens in males so that sperm cannot pass through.

Teacher's Note:
a) Do not confuse fallopian tube (site of fertilisation) with uterus (site of implantation).
b) Vasectomy in males corresponds to tubectomy in females.

OR

(B) (i) Differentiate between self-pollination and cross-pollination.
(ii) Identify A, B and C in the diagram given below and write one function of each. [5 Marks]

[Figure: Diagram of a flower's pistil showing part A pointing to the top structures (two stigmas with pollen grains), part B pointing to a tube-like structure going down through the style, and part C pointing to a structure inside the ovary]

Answer:
1. Self-pollination is the transfer of pollen grains from the anther to the stigma of the same flower, while cross-pollination is the transfer of pollen grains from the anther of one flower to the stigma of another flower of the same species.
2. A - Stigma: Receives pollen grains and provides a suitable environment for their germination.
3. B - Pollen tube: Carries the male germ cells (gametes) down to the female gamete situated in the ovary.
4. C - Egg cell (female germ cell): Fuses with the male gamete to form the zygote.

Teacher's Note:
a) Self-pollination occurs within one flower; cross-pollination occurs between two different flowers of the same species.
b) Fusion of male and female gametes to form zygote is called fertilisation.

36. (A) (i) The power of a lens 'X' is - 2.5 D. Name the lens and determine its focal length in cm. For which eye defect of vision will an optician prescribe this type of lens as a corrective lens?
(ii) "The value of magnification 'm' for a lens is - 2." Using new Cartesian Sign Convention and considering that an object is placed at a distance of 20 cm from the optical centre of this lens, state:
(I) the nature of the image formed;
(II) size of the image compared to the size of the object;
(III) position of the image, and
(IV) sign of the height of the image.
(iii) The numerical values of the focal lengths of two lenses A and B are 10 cm and 20 cm respectively. Which one of the two will show higher degree of convergence/divergence? Give reason to justify your answer. [5 Marks]

Answer:
1. Since power is negative, lens X is a concave lens; \( f = \frac{1}{P} = \frac{100}{-2.5} = -40\,cm \)
2. This type of lens is prescribed for myopia (near-sightedness).
3. Since magnification m = -2 (negative and greater than 1 in magnitude), the image is real and inverted.
4. The image size is magnified, being double the size of the object.
5. The image is formed beyond 2F, on the opposite side of the lens from the object, and the sign of the height of the image is negative.
6. The lens with focal length 10 cm shows a higher degree of convergence/divergence, because a lens with smaller focal length has greater power (converging or diverging ability).

Teacher's Note:
a) Negative power always indicates a diverging (concave) lens; positive power indicates converging (convex) lens.
b) Power is inversely proportional to focal length, so smaller focal length means higher power.

OR

(B) (i) Draw a ray diagram to show the refraction of a ray of light through a rectangular glass slab when it falls obliquely from air into glass.
(ii) State Snell's law of refraction of light.
(iii) Differentiate between the virtual images formed by a convex lens and a concave lens on the basis of:
(I) object distance, and
(II) magnification. [5 Marks]

Answer: (i) When a ray of light travels obliquely from air into a denser medium like glass, it bends towards the normal at the point of entry; on emerging from the glass into air on the other parallel face, it bends away from the normal and emerges parallel to the original incident ray, but laterally displaced.
(ii) Snell's law states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a given pair of media and for light of a given colour: \( \frac{\sin i}{\sin r} = constant \)
(iii) For a convex lens, a virtual image is formed only when the object is placed between the optical centre and the focus, whereas for a concave lens, a virtual image is formed for the object placed anywhere in front of the lens. Also, a convex lens gives a magnified virtual image, while a concave lens always gives a diminished virtual image.

Teacher's Note:
a) Marks are deducted if the direction arrows on the ray diagram are missing.
b) Remember: convex lens forms both real and virtual images depending on object position, but concave lens forms only virtual images.

SECTION E

37. Seawater contains many salts dissolved in it. Common salt is separated from these salts. Deposits of solid salt are also found in several parts of the world. These large crystals are often brown due to impurities. This is called rock salt and is mined like coal. The common salt is an important raw material for chemicals of daily use.

(a) Write balanced chemical equations to show the products formed during electrolysis of brine. [1 Mark]

Answer: \( 2NaCl + 2H_2O \xrightarrow{electricity} 2NaOH + H_2 + Cl_2 \)

Teacher's Note:
a) This process is called the chlor-alkali process.
b) The three products are sodium hydroxide, hydrogen gas, and chlorine gas.

(b) List two uses of any one product obtained during electrolysis of brine. [1 Mark]

Answer: Uses of NaOH: used in degreasing metals, and in making soaps and detergents.

Teacher's Note:
a) Alternative answers about uses of H\(_2\) (as fuel, for making ammonia) or Cl\(_2\) (in water treatment, making bleach) are also accepted.
b) Choose only one product and give exactly two of its uses.

(c) (i) A mild non-corrosive basic salt 'A', used for faster cooking, is strongly heated to produce a compound 'B', that is used for removing permanent hardness of water. Identify A and B and also write the equation for the reaction that occurs when A is heated. [2 Marks]

Answer:
1. A = Sodium hydrogen carbonate (baking soda, NaHCO\(_3\)); B = Sodium carbonate (Na\(_2\)CO\(_3\))
2. \( 2NaHCO_3 \xrightarrow{heat} Na_2CO_3 + H_2O + CO_2 \)

Teacher's Note:
a) Baking soda is used in cooking to make food soft and fluffy.
b) Washing soda (obtained here as B) is used to remove permanent hardness of water.

OR

(c) (ii) Define water of crystallisation. Give two examples of salts that have water of crystallisation. [2 Marks]

Answer:
1. Water of crystallisation is the fixed number of water molecules present in one formula unit of a salt.
2. Examples: Copper sulphate pentahydrate (CuSO\(_4\).5H\(_2\)O) and Gypsum (CaSO\(_4\).2H\(_2\)O).

Teacher's Note:
a) Other accepted examples include washing soda (Na\(_2\)CO\(_3\).10H\(_2\)O) and green vitriol (FeSO\(_4\).7H\(_2\)O).
b) Heating removes water of crystallisation, often changing the colour of the salt (e.g., blue CuSO\(_4\).5H\(_2\)O turns white).

38. The maintenance functions of all living organisms must go on even when they are not doing anything particular. Even when we are just sitting in a class or even asleep, this maintenance job has to go on. These maintenance processes require energy to prevent damage and break-down of cells and tissues, which is obtained by the individual organism from the food prepared by the autotrophs, called producers.

(a) Name and define the process by which green plants prepare food. [1 Mark]

Answer: Photosynthesis is the process by which green plants capture sunlight and convert carbon dioxide and water into carbohydrates with the help of chlorophyll.

Teacher's Note:
a) The key raw materials are carbon dioxide, water, and sunlight, with chlorophyll acting as a catalyst-like pigment.
b) Oxygen is released as a by-product of this process.

(b) Write chemical equation involved in the above process. [1 Mark]

Answer: \( 6CO_2 + 12H_2O \xrightarrow[Sunlight]{Chlorophyll} C_6H_{12}O_6 + 6O_2 + 6H_2O \)

Teacher's Note:
a) Mention both chlorophyll and sunlight as conditions above the arrow for full marks.
b) Glucose (C\(_6\)H\(_{12}\)O\(_6\)) is the main food product formed.

(c) (i) State in proper sequence the events that occur in synthesis of food by desert plants. [2 Marks]

Answer:
1. Absorption of light energy by chlorophyll.
2. Conversion of light energy into chemical energy, and splitting of water into hydrogen and oxygen.
3. Reduction of carbon dioxide to carbohydrates.
4. Desert plants take up CO\(_2\) at night, storing it as an intermediate compound, which is then acted upon using the energy absorbed by chlorophyll during the day.

Teacher's Note:
a) This adaptation in desert plants helps them conserve water by keeping stomata closed during the hot day.
b) The correct sequence, especially the night-time CO\(_2\) uptake step, is the key value point.

OR

(c) (ii) Explain giving reasons what happens to the rate at which the green plants will prepare food
(I) during cloudy weather, and
(II) when stomata get blocked due to dust. [2 Marks]

Answer:
1. During cloudy weather, the rate of photosynthesis decreases due to a low amount of available sunlight.
2. When stomata are blocked by dust, the rate of photosynthesis decreases because gaseous exchange (uptake of CO\(_2\) and release of O\(_2\)) is reduced.

Teacher's Note:
a) Sunlight and gaseous exchange are both limiting factors for photosynthesis.
b) Link each condition to the specific factor of photosynthesis it affects for full marks.

39. In our homes, we receive the supply of electric power through a main supply also called mains, either supported through overhead electric poles or by underground cables. In our country the potential difference between the two wires (live wire and neutral wire) of this supply is 220 V.

(a) Write the colours of the insulation covers of the line wires through which supply comes to our homes. [1 Mark]

Answer: The live wire has red insulation, and the neutral wire has black insulation.

Teacher's Note:
a) The earth wire is usually given green insulation, though it is not directly asked here.
b) Knowing wire colours is important for basic electrical safety.

(b) What should be the current rating of the electric circuit (220 V) so that an electric iron of 1 kW power rating can be operated? [1 Mark]

Answer:
1. Given: P = 1 kW = 1000 W, V = 220 V
2. \( I = \frac{P}{V} = \frac{1000}{220} = 4.54\,A \)
3. The current rating of the circuit should be 5 A.

Teacher's Note:
a) Always round up the calculated current to the next standard fuse/circuit rating for safety.
b) Common standard current ratings available are 5A and 15A circuits in homes.

(c) (i) What is the function of the earth wire? State the advantage of the earth wire in domestic electric appliances such as electric iron. [2 Marks]

Answer:
1. The earth wire provides a low resistance conducting path for current, ensuring that any current leaking to the metallic body of the appliance flows safely to the earth, keeping the body's potential the same as that of the earth.
2. This ensures that the user does not get an electric shock when touching the metallic body of the appliance.

Teacher's Note:
a) The earth wire is a crucial safety feature connected to the metal casing of appliances.
b) This is different from the neutral wire, which completes the normal circuit.

OR

(c) (ii) List two precautions to be taken to avoid electrical accidents. State how these precautions prevent possible damage to the circuit/appliance. [2 Marks]

Answer:
1. Using a fuse wire in the circuit prevents damage to the circuit due to overloading or short-circuiting, as it melts and breaks the circuit when current exceeds a safe limit.
2. Using an earth wire prevents electric shock due to leakage of current, by safely conducting the leaked current to the ground.

Teacher's Note:
a) A fuse protects the circuit and appliance; an earth wire protects the user from shock.
b) Both these safety devices work together to prevent electrical accidents at home.

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